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DerKrebs [107]
3 years ago
7

If the theoretical yield of a reaction is 0.132 g and the actual yield is 0.131 g , what is the percent yield?

Chemistry
1 answer:
Zepler [3.9K]3 years ago
6 0
I believe the percent yield is 99.24%.
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A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pH when 25.00 mL of oxalic acid solution i
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Answer:

pH = 4.09

Explanation:

molarity of oxalic acid in the solution

         = 0.1 x 25 / (25 + 35)

         = 0.0417 M

molarity of NaOH in the solution

          = 0.1 x 35 / (25 +35)

         = 0.0583 M

H2C2O4 + NaOH -------------------> NaHC2O4 + H2O

0.0417          0.0583                            0                0

0                      0.0166                         0.0417

now second acid -base titration  

NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O

0.0417             0.0166                              0                0

0.0251                 0                                  0.0166         ---

now

pH = pKa2 + log [Na2C2O4 / NaHC2O4]

pH = 4.27 + log (0.0166 / 0.0251)

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3 0
2 years ago
1. Using the balanced equation, answer the following questions:
nalin [4]

Answer:

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

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Answer:

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Explanation:

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